Solve Wave Equation from Scratch [2013 HSSP]

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1 1 Solve Wave Equation from Scratch [2013 HSSP] Yuqi Zhu MIT Department of Physics, 77 Massachusetts Ave., Cambridge, MA (Dated: August 18, 2013) I. COURSE INFO Topics Date 07/07 Comple number, Cauchy-Riemann Equation, Cauchy Theorem 14/07 Cauchy Theorem, Cauchy Integral Formula, ODE 21/07 Cauchy integral Theorem, Fourier Transform 28/07 Fourier Series, Fourier Transform, 1st order PDE 04/08 1st order PDE, Laplace Transform 11/08 Operator, ODE, Green s Function 18/08 Green s function of Wave Equation, Solution to Wave Equation Contact HSSP Phone: (617) summer-hssp@mit.edu YZ Phone: no please yqzhu@mit.edu II. Q & A 07/07 1. Do not know much about multivariable calculus -This course only involves the very basics of multivariable calculus, such as partial derivative; there won t be Divergence theorem, Green s theorem etc. There are lots of materials accessible on the Internet for multivariable calculus, such as MIT s OCW and Youtube videos. Besides, ask questions during lectures whenever it s not clear to you. 14/07 2. Don t understand Double pole -See the first two sections (A and B) of notes. 21/07 3. Why do some integrals go to zero? There are two ways we can conclude an integral is equal to zero: using Cauchy theorem and not using Cauchy theorem. For the second case, there are two possibilities: 1. Very large contour - change to polar coordinates and take R to infinity; if the power of R is larger in the denominator the integral is zero; 2. Very small contour - change to polar coordinates and take R to 0; if the power of R is larger in the numerator the integral is zero I presume you are complaining about those ones in the eample of branch cuts. In that case you have a very small contour. The power of R is 1 in the numerator, the power of R is 1/2 in the denominator because 1+ and 1- are 1 The materials of this document will be cumulative because I don t want to create separate Late files and you shouldn t complain.

2 Solve Wave Equation from Scratch [2013 HSSP] 2 both approimated to 1. If it is instead of 1+ the integral will blow up. 4. Branch cuts and branch points Branch points are points that don t have unique values due to rotations. The purpose of branch cuts is to prevent ambiguities; branch cuts are in walls that contours cannot cross. If the rotation is never completed or the contour is never closed, there won t be any ambiguity. With the purpose of branch cuts in mind, it might be easier to think of how to make them. Branch cut can be finite or infinite as long as it serves its purpose. If there s only one branch point, it is always infinity; if there are two or three branch points the branch cut might be finite. However, we can still close the contour if there is an infinite branch point. Regrading Cauchy integrals, the integrals that enclose branch points are not well defined, so it cannot be evaluated; we must find a closed contour that doesn t have any branch point. 5. Fourier Transform We will talk more about it net time. See below (D) for physics-related topics. 28/07 6. Proof of Inversion theorem. We assume that inversion theorem of Fourier transform holds if we can get the same function back using inversion. It is fair as a mathematical assumption (maybe not as a philosophical assumption). There are a few steps in the proof of inversion theorem: change the limits of integral from infinities to finite end points, apply the inversion theorem to f(k), apply the Fourier transform to f() and take the finite end points to infinities. Following this scheme, you should be able to prove the inverse transform. 7. Regarding Cauchy integral and Fourier transform: 1. Integrating in comple plane enables you to evaluate integrals (by hand) which you cannot do or have a very hard time doing using calculus. 2. When doing Fourier transform whose nature is an integral, you might encounter integrals that is easy to evaluate in comple plane and difficult in calculus. 3. With the help of Mathematica or other softwares, you can evaluate almost all integrals without knowing Cauchy theorem, but almost all is not all, I ve done, in my problem sets for this class, many problems (set up by a skilful instructor) that are not solved by Mathematica. So I think it is still good to know; it s a nice eercise for brain as well. 8. Fourier Series [I am not sure if I understand your questions] The idea of Fourier series is to write a function in terms of sines and cosines or equivalently comple eponentials. Only differentiable functions with finite end points can be written in this form; a countereample is e. Fortunately, most functions that are encountered in science and engineering have such properties. All the definitions are quite straightforward. I guess the confusing part is the differential operator story; we might as well discuss it in class if time permits. 9. Partial Differential Equation We will continue net time. 10. Convolution Convolution is an operation for two or more functions. We will talk more about it when it comes to Laplace transform. It is also very common in probability theory, where you might want to convolve two probability distributions to obtain the total probability distribution. 04/ Partial differential equation s Characteristic curves Along the characteristic curves, the function either is constant for homogeneous PDE, or has a constant rate of change for inhomogeneous rate of change. 12. What is inversion theorem of Laplace transform? The function f(), for eample, is the description of an object in space; the Laplace transform of it is the description of the SAME object in a different language. The reason to perform Laplace, which is the also the reason to apply any transform, is that some calculation is simpler in one space than the other. But after you obtain an epression in terms of s, you might want to transform it back, into a space where you can interoperate more easily, or so to speak, into a language which you are more familiar with. Then you need to use the inversion theorem. 13. Connection ODE and Laplace Transform Laplace transform is generic method of solving Ordinary Differential Equations. The reason to perform Laplace,

3 Solve Wave Equation from Scratch [2013 HSSP] 3 which is the also the reason to apply any transform, is to simplify the calculation. The Laplace transform of derivatives can be related to the Laplace transform of the function itself (through integration by parting); this allows you to calculate the eplicit epress of the Laplace transform of the function, from which you can obtain the solution by applying inversion theorem. 11/ Definition of Green s function We haven t got to the point of defining Green s function (but it is always a good idea to know the definition), so I will say more on it net time. It is the function that yields a Dirac delta function when the PDE operator acts on in. For Wave Equation, for eample, the operator is t 2, Green s function is the function such that, ( t 2 )G( ) = δ (4) ( )] 15. Delta function I always refer to Dirac Delta when I say Delta function. It has infinitesimal width and infinite height, but its integral is one , Laplace or Laplacian I found out that Wikipedia calls it Laplace operator or Laplacian and Wolfram calls it Laplacian. 18/ What is G? The Green s Function 18.U( k, t) into formula? Plug U( k, t) into the Wave Equation and you should obtain: ( k t )U( k, t) = 0. We did this step last week; refer to your note for detail. 19. G 1, G 2 substitution In order to obtain the solution of wave equation, we substitute the Green s functions into: u(, t) = d n [G 1 ( )a( ) + G 2 ( )b( )] Before substituting, make sure you have the right Green s function. 20. Motion in String Theory A string is still a string in String Theory, so it moves like a string. However, in order to have a relativistic string, we need to incorporate time coordinate. III. SUPPLEMENTARY NOTES Chances are there will be typos and mistakes since the notes are created by me. Please let me know if there s anything wrong or confusing. A. Taylor Series and Laurent Series The idea of Taylor series is that all analytic functions (i.e. functions whose nth derivative eists) can be epressed as the sum of polynomials. The coefficients are determined by the derivative at a particular point. The function is written as, f(z) = f(z 0 ) + f (z 0 ) 1! To save some ink, we rename the coefficients, (z z 0 ) + f (z 0 ) 2! (z z 0 ) f (n) (z 0 ) (z z 0 ) n +... (1) n! f(z) = a 0 + a 1 (z z 0 ) + a 2 z z 0 ) a n (z z 0 ) n +... (2)

4 Solve Wave Equation from Scratch [2013 HSSP] 4 When the function is no longer analytic, it cannot be epressed as the sum of nice polynomials because it has singularities. So we think of etending Taylor series into a more generic power series, namely, Laurent series. If f(z) has a singularity at z 0 and it is a simple pole, then f(z) can be epressed as, f(z) = If f(z) has a double pole at z 0, f(z) = a 1 z z 0 + a 0 + a 1 (z z 0 ) + a 2 z z 0 ) a n (z z 0 ) n +... (3) a 2 (z z 0 ) 2 + a 1 z z 0 + a 0 + a 1 (z z 0 ) + a 2 (z z 0 ) a n (z z 0 ) n +... (4) B. Residual and Double Pole Residual is simple a 1 of a Laurent series. From this definition, we know that it depends on the function, the contour and the singularities enclosed. In order to find a 1, we multiply both sides by (z z 0 ) and take the limit of z goes to z 0, lim f(z)(z z 0 ) = lim (a 1 + a 0 (z z 0 ) + a 1 (z z 0 ) 2 + a 2 (z z 0 ) a n (z z 0 ) n ) (5) z z 0 z z 0 This is what happens when we evaluate the integral. When z 0 is a double pole, the same trick doesn t work because of the a 2 term, lim f(z)(z z 0 ) = lim (fraca 2 z z 0 + a 1 + a 0 (z z 0 ) + a 1 (z z 0 ) 2 + a 2 (z z 0 ) a n (z z 0 ) n ) (6) z z 0 z z 0 In order to obtain a 1, we do instead, lim z z 0 d dz [f(z)(z z 0) 2 ] = lim z z 0 d dz (a 2 + a 1 (z z 0 ) + a 0 (z z 0 ) 2 + a 1 (z z 0 ) a n (z z 0 ) n ) (7) Clearly, a 1 is the only thing left after this. 1 If a function contains (z z 0) or even worse, we multiply whatever that sends the denominator to 1, and take 3 derivatives. Always keep in mind what you want to obtain before manipulating the function. C. Principle Value of an Integral Principle value method is a way to evaluate integral. It is neither smatter nor simpler than Cauchy integral formula in my opinion. I was taught this because I might need to communicate with the rest of world. The difference before principle value and Cauchy method is that there is an etra step in principle value before applying Cauchy integral formula. The idea of principle value is that we jump over the singularities when evaluating an integral. This etra step which presumably makes it more straightforward. Principle value of an integral is the same as the integral s value when the integral converges, yet it eists even if the integral diverges (i.e. goes to infinity). P = lim ɛ The integral from ɛ to ɛ goes to zero as ɛ goes to zero. Eample: I = f(z)dz + lim sin d ɛ f(z)dz (8) In this eample I will only demonstrate the etra step of Principle value method to show how principle value method is connected to Cauchy theorem. The rest of the eample which is the Cauchy method will be completed in class. Solution:

5 Solve Wave Equation from Scratch [2013 HSSP] 5 First note, the principle value of this integral is the true value of the integral because the function is convergent. There is a singularity at 0, but the function doesn t go to infinity there. So, I = P sin d. sin is the imaginary part of ei by Euler s equation. Now define J = P e i d and I = Im(J). ɛ J = lim e i e i d + lim ɛ d This brings us to the starting point of Cauchy theorem. D. Fourier Transform Heisenberg uncertainty principle (position and momentum): p > h/2 (there is also an energy and time uncertainty principle). It says momentum and position of an object cannot be determined with 100% certainty simultaneously. Mathematically, the Fourier Transform of Dirac delta is a sinusoidal curve. Using physical intuition, we know this makes sense because when position is very well defined (Dirac delta), its momentum spreads out. This is one eample showing math and physics agree with each other. f() and f(k) describe the same object but in two different languages. An interesting subtlety concerning Fourier transform is that nothing is infinity is physical world, so f() and f(k) are not equivalent unless they are infinite. This somehow seconds that transform is not the same as a change of variable. However, it is hard to think at infinity; this is easier to see if you think of and p as matrices. In non-commutative algebra (whatever that means), Heisenberg principle is [, p] = i hi. The dimensions of trace are not equal on the two sides unless and p have infinite dimensions. tl; dr (totally your loss if you didn t.) Math goes hand in hand with physics. E. Fourier series Applet: [Courtesy of MIT instructors] F. Useful Tools Late - Allows you to type everything and anything; Easy to format; used in (almost) all science and math journals; Free MATLAB - Good for matri and differential equations (yet not the best and sometimes crashes); C++ synta; Popular for data analysis (popular is not necessarily the best); not free Mathematica - Good for symbolic manipulations; Creates beautiful graphs; has its own synta, but it s super straightforward; Help Centre actually helps; not free MATLAB and Mathematica are not free, actually quite pricey, but chances are your college will pay for you. They doesn t help in either high school math or Olympiad. I didn t use any of these in high school, but they are etremely helpful to me in college.